# Schwarz Lemma/Conformal mapping problem

Let $F:\mathbb{H}\rightarrow \mathbb{D}$ be holomorphic, where $\mathbb{H}$ is the upper half plane and $\mathbb{D}$ is the unit disc. Show that if $F(i)=0$, then $$|F(z)|\leq \left|\frac{z-i}{z+i}\right|$$ for all $z\in\mathbb{H}$.

I have trouble constructing an auxiliary function to apply the Schwarz Lemma or something, any hints?

Thanks

• You need a conformal $T\colon \mathbb{D}\to \mathbb{H}$ with $T(0) = i$. – Daniel Fischer Dec 15 '14 at 2:16
• Look at the given right hand side for inspiration about how to construct the $T$ in Daniel's hint. – mrf Dec 15 '14 at 7:59

Consider $G(z)=\dfrac{z-i}{z+1}$ and $g(z)=F\circ G^{-1}(z)$. We have $g(0)=F\circ G^{-1}(0)=F(i)=0$. By Schwarz's Lemma, $|g(z)|\leqslant |z|$ so $|F\circ G^{-1}(z)|\leqslant |z|$ and then $|F(z)|\leqslant |G(z)|=\left|\dfrac{z-i}{z+i}\right|$
The conformal mapping from the unit disk to the upper half-plane is $G=\frac{z-1}{z+1}$. So, $F\circ G$ will satisfy the hypotheses of the Schwarz Lemma.